In the drawer lies some. Tasks for the classical definition of probability. Examples of solutions
In the box lies some of the white and black balls. If you randomly pull out from there two balls, then the likelihood that they are both will be white, equal to 1/2.
but) What Minimal possible number of balls in the box?
b) The same question is provided that the black balls are a self-number.
Tip 1.
Suppose in the box contain w. White balls I. b. Black balls. What is the likelihood that if you randomly pull out two balls from the box, then they both will be white?
Tip 2.
Try to find the desired value. w. For small values b. (eg, b.= 1, b.= 2, b.= 3, ...).
Decision
So, let the box contain w. White balls I. b. Black balls. For simplicity, we assume that we pull the balls from the box sequentially. Then the likelihood that the first ball that we extracted from the box will be white, equal to \\ (\\ FRAC (W) (W + B) \\), and the likelihood that the second ball also turns out to be white (provided that the first ball White) is \\ (\\ FRAC (W-1) (W + B-1) \\). Under the condition of the task, the likelihood that both balls are white, equal to 1/2, that is
\\ [\\ dfrac (w) (W + B) \\ CDOT \\ DFRAC (W-1) (W + B - 1) \u003d \\ DFRAC12 \\ QQuad \\ Qquad \\ Qquad \\ Qquad (1) \\]
Note that the resulting formula will not change if we assume that the balls are retrieved from the box at the same time. In fact, the number of opportunities to pull out two arbitrary balls equals
\\ (C_ (W + B) ^ 2 \u003d \\ DFRAC ((W + B) (W + B-1)) (2) \\),
get two white balls
\\ (C_W ^ 2 \u003d \\ DFRAC (W (W-1)) (2) \\) methods.
That is, the desired probability is equal
\\ (\\ dfrac (C_W ^ 2) (C_ (W + B) ^ 2) \u003d \\ FRAC (W) (W + B) \\ CDOT \\ DFRAC (W-1) (W + B - 1) \\).
The resulting expression can be considered as an equation from two variables: w. and b.. Considering that we need to find the smallest value of the expression ( w. + b.), It is most natural to try to solve this equation of prosperity. It is, it is logical to try to consistently substitute the values b. = 1, b. = 2, b. \u003d 3, ..., and then find out whether the resulting relative w. quadratic equation whole solutions or not. In our case, this method pretty quickly leads to a solution. So, for b. \u003d 1 We also get a linear equation:
\\ [\\ DFRAC (w (W-1)) (w (w + 1)) \u003d \\ dfrac12 \\ qquad \\ rightarrow \\ qquad 2 (W-1) \u003d W + 1 \\ Qquad \\ leftrightarrow \\ qquad w \u003d 3 \\]
This gives us a decision of item a). The same method, tickling a bit, one could find an answer for items b), but we will go to another, more elegant from a mathematical point of view, by.
Note that b. \u003e 0 I. w. \u003e 0 inequality
\\ [\\ DFRAC (W) (W + B)\u003e \\ DFRAC (W-1) (W + B-1). \\]
Taking into account the equation (1) hence it follows that
\\ [\\ left (\\ dfrac (w) (W + B) \\ Right) ^ 2\u003e \\ dfrac12\u003e \\ left (\\ dfrac (W-1) (W + B-1) \\ Right) ^ 2. \\]
Removing square roots, have
\\ [\\ dfrac (w) (W + B)\u003e \\ DFRAC (1) (\\ SQRT2)\u003e \\ DFRAC (W-1) (W + B-1). \\]
Consider separately the first of these inequalities:
\\ [\\ dfrac (w) (W + B)\u003e \\ DFRAC (1) (\\ sqrt2) \\ qquad \\ rightarrow \\ qquad w \\ sqrt2\u003e w + b \\ qquad \\ rightarrow \\ qquad w\u003e \\ dfrac (b) (\\ Similarly, for the second inequality we have
\\ [\\ dfrac (1) (\\ sqrt2)\u003e \\ DFRAC (W-1) (W + B-1) \\ Qquad \\ rightarrow \\ qquad w + b + 1\u003e (W-1) \\ sqrt2 \\ qquad \\ rightarrow \\ Thus, we gain an assessment for
Through the magnitude<\dfrac{b+\sqrt2-1}{\sqrt2-1}=(\sqrt2+1)b+1.\]
\\ [1+ \\ SQRT2) B + 1\u003e W\u003e (1+ \\ SQRT2) B. \\] w. For example, for b.:
\u003d 1, considering that \\ (1 (,) 414
\u003d 3. In our case (when b. \u003d 1 I.<\sqrt2<1{,}415\), мы получаем неравенство \(2{,}414
\u003d 6, ... and the corresponding values
until we meet the appropriate. b. = 2, b. = 4, b. In the interval w.appropriate
| b. | w. P. | \\ (\\ dfrac57 \\ cdot \\ dfrac46 \\ ne \\ dfrac12 \\) w. | \\ (\\ DFRAC (10) (14) \\ CDOT \\ DFRAC9 (13) \\ NE \\ DFRAC12 \\) |
| 2 | (4,82; 5,83) | 5 | \\ (\\ dfrac (15) (21) \\ CDOT \\ DFRAC (14) (20) \u003d \\ DFRAC12 \\) |
| 4 | (9,65; 10,66) | 10 | Thus, if |
| 6 | (14,48; 15,49) | 15 | Even, the minimum number of balls in the box is 21. |
Afterword b. A naturally emerging the reader after solving the problem is how to find all possible sets of black and white balls, for which the probability of extracting two white balls from the box is 1/2. To do this, consider equation (1) as an equation from a variable
, and magnitude
We will consider the parameter. We rewrite this equation as follows: w.\\ [\\ dfrac (w) (W + B) \\ CDOT \\ DFRAC (W-1) (W + B - 1) \u003d \\ dfrac12 \\ qquad \\ rightarrow \\ qquad 2w ^ 2-2W \u003d W ^ 2 + 2WB + B ^ 2-WB \\ QQuad \\ Rightarrow \\ Qquad \\] \\ b. It is clear that this equation has an integer solution then and only if its discriminant is a square of an integer. In other words, for some integer
M.
Fair equality or, that the same, \\ (m ^ 2-8b ^ 2 \u003d 1 \\). The resulting equation represents a special case of integer equations of a more general form:
here
D.
- A predetermined integer that is not a complete square. Such equations with a light hand of Leonard Euler are traditionally called Pella equations (although English mathematician John Pell, in whose heler and called such equations, most likely nothing has nothing to do with them). It turns out, with each satisfying condition, the value of the parameter The equation of this type has infinitely many solutions, and all these solutions are uniformly.
We will demonstrate on the example what is meant. Let be \u003d 2. Then it is not difficult to show that if a pair ( x., y.) is a solution to the equation \\ (x ^ 2-2Y ^ 2 \u003d 1 \\), then the pair (3 x. + 4y., 2x. + 3y.) They are also. Indeed,
\\ [(3x + 4y) ^ 2-2 (2x + 3Y) ^ 2 \u003d (9x ^ 2 + 24XY + 16Y ^ 2) -2 (4x ^ 2 + 12XY + 9Y ^ 2) \u003d x ^ 2-2Y ^ 2. \\]
Thus, based on the trivial solution (1, 0), we can get an infinite sequence of various solutions \\ ((x_k, y_k) \\) with the help of a recurrent formula
\\ ((x_k, y_k) \u003d f (x_ (k - 1), y_ (k-1)) \\),
\\ (f (x, y) \u003d (3x + 4y, 2x + 3y) \\).
In our case, solutions are obtained by: (3, 2), (17, 12), (99, 70), (577, 408), ...
It turns out that these values \u200b\u200bare the positive solutions of the equation \\ (x ^ 2-dy ^ 2 \u003d 1 \\) are exhausted, and its remaining solutions differ from those indicated only by the sign.
Similarly, things are also in the general case. Interest here represents several important points. First, all non-trivial positive solutions can be obtained by multiple multiplication of one of them, which we will call basic, to myself. Under the "multiplication" of two solutions of the Pella equation, we mean the next tricky operation (which is the habit as well as the usual multiplication - point):
\\ [(x_1, y_1) \\ cdot (x_2, y_2) \u003d (x_1x_2 + dy_1y_2, x_1y_2 + x_2y_1). \\]
\u003d 3. In our case (when \u003d 2 The main solution is (3, 2), and the multiplication of an arbitrary solution has a view
\\ ((x, y) \\ cdot (3.2) \u003d (3x + 4y, 2x + 3y) \\) - this is exactly what we talked above.
Secondly, to find out how to find the notorious basic solution for each specific value. . Here we suddenly help chain fractions: it turns out, any positive decision ( x., y.) The Pelly equations correspond to the appropriate fraction \\ (\\ FRAC (X) (Y) \\) of the number \\ (\\ sqrt (d) \\). However, the opposite statement is incorrect: not every suitable fraction corresponds to the solution of the Pell equation, but only those whose numbers are ( kN.- 1) where n. - The length of the sequence of chain fraction elements for the number \\ (\\ sqrt (d) \\). For some These numbers (and, ~ as a result, positive solutions) may be very large. So, for \u003d 61 The main solution has the form (1 766 319 049, 226 153 980).
Finally, it will be curious that in the proof of the existence of a nontrivial solution of the Pella equation, the geometric lemma of Minkowski is plays a key role about the convex body. This lemma unexpectedly arises in a variety of tasks of the theory of numbers and is the brightest example of the connection of algebra and geometry in higher mathematics.
In the preparation of the article, the following materials were used:
1) V. O. Bugaenko. "Pella Equations" (Library "Mathematical Enlightenment", Issue 13).
2) F. MOSTELLER. "Fifty entertaining probabilistic tasks with solutions."
3) Articles V. Sanderov and A. Spivak about the Pelly equation in the journal "Kvant" (
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Famous corporations - Google, Intel or Apple - are famous for the fact that there are caverzy challenges in the interview. Site ain.ua collected 10 interesting examples of such tasks. Some of them offered the company themselves, and some have posted users who have already held interviews. To solve them require knowledge of mathematics at the school level or simply an email.
website Offers check how you coped with such an interview.
What is asked in apple
Task 1.
The task of logic. Sheldon Cooper (the very brilliant physicist from the popular series) reached the challenge quests in pursuit of treasures until the last turn. In front of him - two doors, one leads to the treasure, the second to the deadly labyrinth. Each door has a guard, each of them knows which door leads to a treasure. One of the guards will never lie, the other is always lying. Sheldon does not know who of them lie, and who is not. Before choosing the door, you can set only one question and only one guard.
Question: What to ask Sheldon at the guard to get to the treasure?
You can ask anyone, at the same time ask a question like this: "What door, according to another guard, is correct?". If he asks at the truthful, then will receive the data on which door leads to the labyrinth, because the guard always lies. If he asks in Handing the guard, again, he learns what the door leads to the labyrinth, because the one-stop on the door will indicate the truthful guard.
Task 2.
The land captured aliens. They plan to destroy the entire planet, but decided to give humanity a chance. They chose a dozen of the smartest people and placed them in a completely dark room, putting in a row, one by one. On each of the people we had a hat, hats of only two colors - pink and green. After all hats turn out to be on their heads, the light turns on.
The aliens begins with the last person in a row and asks about what color hat is on his head. Other words, except for the color of the hat, can not be pronounced. To relent - too. If he responds true, remains alive, mistakes - he is killed.
You can not see what color your hat is, but you can agree on a certain principle for which to respond to everyone. The arrangement of the hats is random, combinations can be anywhere, you can see all hats that are located in front of you.
Question: What needs to be responsible to survive as many people as possible?
The first responding considers the number of green hats in front of him: if it is an odd number, he calls "green", if even - "pink". The next, seeing the amount and color of the hats in front of them, can thus calculate how the hat is on his head (for example, if the green is still an odd amount, it is obvious that it is pink), and so on. Thus, it is guaranteed to survive 9 out of 10, and the first one who answered the chance 1 to 1.
What is asked in Adobe
Task 3.
You have 50 motorcycle filled with fuel tank, which is enough for 100 km drive.
Question: Using these 50 motorcycles, how far can you come out (considering that initially they are in the conditionally one point of space)?
The easiest answer: to have them all at the same time and drive 100 km. But there is another solution. First, move all the motorcycles by 50 km. Then prohibit fuel from half of the motorcycles to another half. In this way, you have 25 motorcycles with full tank. Drive another 50 km and repeat the procedure. So you can climb 350 km (without taking into account that fuel, which will remain from the "extra" motorcycle in Section 25 in Num).
What is asked in Microsoft
Task 4.
You have an endless stock of water and two buckets - on 5 liters and 3 liters.
Question: How would you measure 4 liters?
Fill the five-liter bucket with water and pour part of the water into three-liter. You have now 3 liters in a small bucket and 2 - in the big one. Empty a small bucket and reclose the remaining 2 liters from the big one. Refill a big bucket and break the water into small from it. There there are already 2 liters of water, so the liter will have to fasten, and 4 liters will remain.
Task 5.
You have two cuts of the rope. Everything is such that if you set fire to it from one end, it will burn exactly 60 minutes.
Question: Having only a box of matches, how to measure with two segments of such a rope of 45 minutes (we can not tear ropes)?
One of the segments is mounted on two ends, at the same time the second segment is ignited, but from one end. When the first cut is completely complete, 30 minutes will pass, the first will also remain a 30-minute segment. Fingering it from two ends, we get 15 minutes.
What ask google
Task 6.
You have 8 balls of the same species and size.
Question: How to find a heavier ball using scales and only two weighing?
Select 6 balls, divide them into groups of 3 balls and put on the scales. The group with a heavier ball will turn the bowl. Choose any 2 balls from this triple and weigh. If a heavy ball among them, you will learn it; If they weigh the same - the hard one that remains. If a heavier ball in groups of 3 balls did not turn out, he is among the 2 remaining.
What is asked in Qualcomm
Task 7.
This task was described by the user whom was interviewed by the position of Senior Systems Engineer. He noted in the description of the task that he had his answer, about which he argued for a long time with a person who conducted an interview.
Suppose we have 10 packet data transmission on a wireless network. The channel is not very high quality, so there is a probability of 1/10 that the data packet will not be transmitted. Transmitter always knows, successfully or unsuccessfully passed the data packet. When the transmission is unsuccessful, the transmitter will transmit the packet until it succeeds.
Question: What channel bandwidth get?
According to the user, the answer should have been: 9 packages per second. But the man who held an interview, did not agree with him, however, did not name the answer, but he repeated that "due to repentransmission, bandwidth should be reduced more than 1/10."
There is a morphological dictionary of approximately 100,000 inputs in which the verbs of the perfect and imperfect species are placed in separate articles (that is, "do" and "do" are considered different vocabulary inputs). You need to find in the dictionary such species pairs and "glue" articles in one.
Question: Describe a general scenario of solving such a task and an exemplary viewing algorithm.
Replies to the tasks of "Yandex" with us, unfortunately, no.
And bonus
Task 10.
This task is attributed to Albert Einstein - allegedly with her help, he picked up assistants. Another almost legendary story attributes to the authorship of Lewis Carroll. Note that it is very simply solved on paper, but if you want hardcore - try to decide in the mind.
There are five houses on the street.
The Englishman lives in a red house.
The Spaniard has a dog.
In a green house drink coffee.
Ukrainian drinks tea.
The green house is right right to the right of the White House.
The one who smokes Old Gold breeds snails.
In the yellow house smoke Kool.
In the central house they drink milk.
Norwegian lives in the first house.
A neighbor who smokes Chesterfield holds a fox.
In the house next to the one in which the horse is kept, smoke Kool.
One who smokes Lucky Strike drinks orange juice.
Japanese smoke Parliament.
Norwegian lives next to the blue house.
Each of the houses is painted in a separate color, in every house there lives a representative of a separate nationality, everyone has its own pet, its favorite brand cigarettes and drink.
Question: Who drinks water? Who keeps zebra?
Tasks on the classical probability definition.
Examples of solutions
In the third lesson, we will consider various tasks regarding the direct use of the classical probability definition. To effectively study the materials of this article, I recommend to get acquainted with the basic concepts probability theories and foundations Combinatorics. The task of the classical definition of the likelihood with a probability striving for a unit will be present in your independent / test work on the tervera, so tune in to serious work. You ask what is serious? ... just one primitive formula. Warning from the levity - the thematic tasks are quite diverse, and many of them can easily put in a dead end. In this regard, in addition to the study of the main lesson, try to explore additional tasks on the topic that are in the piggy bank. ready-made solutions for higher mathematics. Receptions of solutions of decisions, and "friends" still "need to know in the face", because even the rich fantasy is limited and typical tasks too. Well, I will try in good quality to disassemble the maximum number.
Remember the classics of the genre:
The likelihood of an event in some test is equal to the ratio, where:
- Total number of all equal possible, elementary outcomes of this test that form full group of events;
- quantity elementary Exodes conducive to events.
And immediately immediate pit stop. Do you understand the underlined terms? Meaning a clear, not intuitive understanding. If not, then it's better to return to the 1st article by probability theories And only after that go on.
Please do not miss the first examples - I will repeat one fundamentally important point, as well as I will tell you how to make a decision and what methods this can be done:
Task 1.
In the urn there are 15 whites, 5 red and 10 black balls. At rags extracted 1 ball, finding the likelihood that it will be: a) white, b) red, c) black.
Decision: The most important prerequisite for the use of the classic probability determination is the ability to calculate the total outcome.
In total in the urn: 15 + 5 + 10 \u003d 30 balls, and obviously, the following facts are valid:
- extracting any ball equally possible (equalityoutcomes), while the outcomes elementary and form full group of events (i.e., as a result of the test, some of the 30 balls will be extracted).
Thus, the total number of outcomes:
Consider an event: - White ball will be removed from the urn. This event is favorable elementary Exodues, therefore, according to the classical definition: 
- The likelihood of the urn will be extracted white ball.
Oddly enough, even in such a simple task you can allow a serious inaccuracy, on which I have already sharpened attention in the first article by probability theories. Where is the underwater stone? It is incorrect to argue that "One half of the balls is white, then the probability of removing a white bowl» . In the classical definition of the probability we are talking about Elementary outcomes and fraction should be prescribed!
With other items, in the same way, consider the following events:
- A red ball will be extracted from the urn;
- Black ball will be extracted from the urn.
Event favors 5 elementary outcomes, and an event - 10 elementary outcomes. Thus, the corresponding probabilities: 
Typical check of many tasks across the terber is carried out using theorems on the sum of the probability of events forming a complete group. In our case, events form a complete group, which means that the sum of the corresponding probabilities should be definitely equal to one :.
Check if it is: what I wanted to make sure.
Answer: 
In principle, the answer can be recorded in more detail, but personally, I used to put there only numbers - for the reason that when you begin to "stamping" tasks with hundreds and thousands, then you strive to reduce the solution as much as possible. By the way, about brevity: in practice, the "high-speed" version of the design is distributed solutions:
Total: 15 + 5 + 10 \u003d 30 balls in urn. By classical definition:
- the likelihood, then the white ball will be extracted from the urn;
- the likelihood of the urn will be extracted with a red ball;
- The likelihood of the urn will be extracted by a black ball.
Answer: 
However, if there are several points in the condition, then the solution is often more convenient to arrange the first way, which takes a little longer time, but it all "folds over the shelves" and makes it easier to navigate in the task.
We warm up:
Task 2.
The store received 30 refrigerators, five of which have a factory defect. Randomly choose one refrigerator. What is the probability that it will be without a defect?
Select the appropriate design option and check the sample at the bottom of the page.
In the simplest examples, the number of common and the amount of favored outcomes lie on the surface, but in most cases the potatoes have to dig independently. The canonical series of tasks about forgetful subscriber:
Task 3.
By dialing the phone number, the subscriber forgot the last two numbers, but remembers that one of them is zero, and the other is odd. Find the likelihood that it will type the correct number.
Note : zero is a different number (divided by 2 without a residue)
Decision: First, find the total number of outcomes. By condition, the subscriber remembers that one of the numbers is zero, and the other digit is odd. Here is more rational not to wise with combinatorics and take advantage method of direct transmission of outcomes
. That is, when making a solution, we simply write down all combinations:
01, 03, 05, 07, 09
10, 30, 50, 70, 90
And count them - total: 10 outcomes.
Favorable outcome one: the right number.
By classical definition:
- the likelihood that the subscriber will score the correct number
Answer: 0,1
Decimal fractions in the theory of probabilities look quite appropriate, but you can adhere to the traditional epismat style, operating only by ordinary fractions.
Advanced task for self solutions:
Task 4.
The subscriber has forgotten the PIN code to his SIM card, however, it remembers that it contains three "five fives", and one of the numbers is "Seven", or the "eight". What is the probability of successful authorization from the first attempt?
Here you can still develop the idea of \u200b\u200bthe likelihood that the subscriber is waiting for Kara in the form of a bunch code, but, unfortunately, the reasoning will already be out of the scope of this lesson
Solution and answer below.
Sometimes the enumeration of combinations is very painful. In particular, the situation in the next, no less popular group of tasks, where 2 playing cubes are thrown out (less often - more):
Task 5.
Find the likelihood that when throwing two playing bones in the amount will fall:
a) five points;
b) no more than four points;
c) from 3 to 9 points inclusive.
Decision: Find the total number of outcomes:
Ways can fall the face of the 1st Cube and ways can fall the edge of the 2nd cube; by rule multiplying combinations, Total: 
Possible combinations. In other words, each The face of the 1st cube can be ordered Couple with each Grand 2nd Cube. We agree to record such a pair in the form where the digit that fell on the 1st cube is a digit that dropped on the 2nd cube. For example:
- on the first cube there were 3 points, on the second - 5 points, the sum of the points: 3 + 5 \u003d 8;
- On the first cube, 6 points fell, on the second - 1 point, the sum of the points: 6 + 1 \u003d 7;
- 2 points fell on both bones, the sum: 2 + 2 \u003d 4.
It is obvious that the smallest amount gives a couple, and the greatest - two "six".
a) Consider the event: - when throwing two playing bones, 5 points will fall. We write and calculate the number of outcomes that favor this event:
Total: 4 favored outcome. By classical definition:
- The desired probability.
b) Consider an event: - no more than 4 points will fall. That is, or 2, or 3, or 4 points. We list again and calculate the favorable combinations, on the left I will record the total number of points, and after the colon - the appropriate pairs: 
Total: 6 conducive combinations. In this way:
- The likelihood that no more than 4 points will fall out.
c) Consider the event: - drops from 3 to 9 points inclusive. Here you can go straight road, but ... something I do not want. Yes, some couples are already listed in previous paragraphs, but the work still has a lot.
How best to do? In such cases, the rational turns out of the Okol Path. Consider opposite event: - 2 or 10 or 11 or 12 points will fall out.
What's the point? The opposite event is favored by a significantly smaller number of pairs: 
Total: 7 favored outcomes.
By classical definition:
- The likelihood that less than three or more than 9 points will fall out.
In addition to direct transfers and counting outcomes, in the go combinatorial formulas. And again the epic task about the elevator:
Task 7.
In the elevator of the 20-storey house on the first floor there were 3 people. And drove. Find the chance that:
a) they will come out on different floors
b) two will come out on the same floor;
c) everyone will come out on the same floor.
Our fascinating occupation approached the end, and finally once again, I strongly recommend that if not to break, then at least understand additional tasks for classic probability definition. As I have already noted, "putting a hand" also matters!
Next by the course - Geometric definition of probability and Theorems of addition and multiplication of probabilities And ... luck in the main thing!
Solutions and answers:
Task 2: Decision: 30 - 5 \u003d 25 refrigerators do not have a defect.
- The likelihood that at random the selected refrigerator does not have a defect.
Answer
:
Task 4: Decision: find the total number of outcomes:
In the ways you can choose the place where the dubious digit is located. and on everyone Of these 4 seats, 2 digits (seed or eight) can be located. According to the rule of multiplication of combinations, the total number of outcomes: 
.
Alternatively, in solving, you can simply list all the outcomes (the benefit of them is a bit):
7555, 8555, 5755, 5855, 5575, 5585, 5557, 5558
The favorable outcome is one (the correct PIN code).
Thus, according to the classical definition:
- the likelihood that the subscriber is authorized from the 1st attempt
Answer
:
Task 6: Decision: Find the total number of outcomes:
In the ways the numbers can fall out on 2 cubes.
a) Consider an event: - When throwing two playing bones, the product of glasses will be equal to seven. For this event, there are no favorable outcomes, according to the classical probability definition:
. This event is impossible.
b) Consider the event: - When throwing two playing bones, the product of points will be at least 20. The following outcomes are conducive to this event:
TOTAL: 8.
By classical definition:
- The desired probability.
c) Consider opposite events:
- The product of glasses will be even;
- The product of glasses will be odd.
We list all the outcomes conducive to events:
Total: 9 favored outcomes.
By classical probability definition: 
The opposite events form a complete group, so:
- The desired probability.
Answer
: 
Task 8: Decision: Calculate the total number of outcomes: 
Ways may fall 10 coins.
Another way: ways can fall 1st coin and ways can fall 2nd coin and … and Ways can fall the 10th coin. According to the rule of multiplication of combinations, 10 coins may fall 
ways.
a) Consider the event: - Orel will fall on all coins. This event is conducive to a single outcome, according to the classical definition of the likelihood :.
b) Consider the event: - Eagle falls on 9 coins, and on one - the rush.
There is coins on which the rush can fall. By classical probability definition: 
.
c) Consider the event: - Eagle falls on half of the coins.
Exists 
Unique combinations of five coins, on which the eagle can fall. By classical probability definition: 
Answer
:
